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Work and Power for Rotation Work = Fd = FRq t = FR q Work = tq tq Work Power = = t t s q w= t F F s = Rq Power = t w Power = Torque x average angular velocity Example 1: The rotating disk has a radius of 40 cm and a mass of 6 kg. Find the work and power if the 2-kg mass is lifted 20 m in 4 s. Work = tq = FR q s 20 m q= = = 50 rad R 0.4 m q 2 kg 6 kg Power = Work = 392 J t 4s F F=W s = 20 m F = mg = (2 kg)(9.8 m/s2); F = 19.6 N Work = (19.6 N)(0.4 m)(50 rad) s Work = 392 J Power = 98 W The Work-Energy Theorem Recall for linear motion that the work done is equal to the change in linear kinetic energy: Fx = ½mv ½mv 2 f 2 0 Using angular analogies, we find the rotational work is equal to the change in rotational kinetic energy: tq = ½Iw ½ Iw 2 f 2 0 Two Kinds of Kinetic Energy Kinetic Energy of Translation: Kinetic Energy of Rotation: K= w ½mv2 v R P K = ½Iw2 Total Kinetic Energy of a Rolling Object: KT = mv I w 1 2 2 1 2 2 Translation or Rotation? If you are to solve for a linear parameter, you must convert all angular terms to linear terms: s q= R v w= R I = (?)mR 2 If you are to solve for an angular parameter, you must convert all linear terms to angular terms: s =qR v = wR Example 2: A circular hoop and a circular disk, each of the same mass and radius, roll at a linear speed v. Compare the kinetic energies. w w Two kinds of energy: KT = ½mv2 v Kr = ½Iw2 Total energy: E = ½mv2 + ½Iw2 2 v 2 2 Disk: E = ½mv ½ ½mR 2 R 2 v 2 2 Hoop: E = ½mv ½ mR 2 R v w= R E = ¾mv2 E = mv2 v Conservation of Energy The total energy is still conserved for systems in rotation and translation. However, rotation must now be considered. Begin: (U + Kt + KR)o = End: (U + Kt + KR)f Height? mgho Rotation? ½Iwo2 velocity? ½mvo2 = mghf Height? ½Iwf2 Rotation? ½mvf2 velocity? Example 3: Find the velocity of the 2-kg mass just before it strikes the floor. R = 50 cm mgho mghf = ½Iwo2 ½Iwf2 ½mvf2 ½mvo2 mgh0 = 12 mv 2 12 I w 2 (2)(9.8)(10) = (2)v (6)v 2 2 kg h = 10 m I = 12 MR 2 2 v 2 2 1 1 1 mgh0 = 2 mv 2 ( 2 MR ) 2 R 1 2 6 kg 1 4 2 2.5v2 = 196 m2/s2 v = 8.85 m/s Example 4: A hoop and a disk roll from the top of an incline. What are their speeds at the bottom if the initial height is 20 m? mgho = ½mv2 + ½Iw2 Hoop: I = mR2 2 v 2 2 mgh0 = ½mv ½(mR ) 2 R 20 m mgho = ½mv2 + ½mv2; mgho = mv2 v = gh0 = (9.8 m/s2 )(20 m) Hoop: Disk: I = ½mR2; mgho = ½mv2 + ½Iw2 2 v 2 2 mgh0 = ½mv ½(½mR ) 2 R v = 14 m/s v= 4 3 gh0 v = 16.2 m/s Angular Momentum Defined Consider a particle m moving with velocity v in a circle of radius r. Define angular momentum L: L = mvr Substituting v= wr, gives: L = m(wr) r = mr2w For extended rotating body: L = (Smr2) w v = wr m w m1 axis m 4 m3 m2 Object rotating at constant w. Since I = Smr2, we have: L = Iw Angular Momentum Example Rank the following from largest to smallest angular momentum. answer Example 5: Find the angular momentum of a thin 4-kg rod of length 2 m if it rotates about its midpoint at a speed of 300 rpm. 1 For rod: I = 12mL2 = 1 (4 12 kg)(2 m)2 L=2m m = 4 kg I = 1.33 kg m2 rev 2 rad 1 min w = 300 = 31.4 rad/s min 1 rev 60 s L = Iw = (1.33 kg m2)(31.4 rad/s)2 L = 1315 kg m2/s Impulse and Momentum Recall for linear motion the linear impulse is equal to the change in linear momentum: F t = mv f mv0 Using angular analogies, we find angular impulse to be equal to the change in angular momentum: t t = Iw f Iw 0 Example 6: A sharp force of 200 N is applied to the edge of a wheel free to rotate. The force acts for 0.002 s. What is the final angular velocity? I = mR2 = (2 kg)(0.4 m)2 I = 0.32 kg m2 Applied torque t = FR t = 0.002 s w = 0 rad/s w o R R = 0.40 m F 2 kg F = 200 N Impulse = change in angular momentum 0 t t = Iwf Iwo FR t = Iwf wf = 0.5 rad/s Conservation of Momentum If no net external torques act on a system then the system’s angular momentum, L, remains constant. Speed and radius can change just as long as angular momentum is constant. 0 Ifwf = Iowo Ifwf Iowo = t t Io = 2 kg m2; wo = 600 rpm I 0w 0 (2 kg m )(600 rpm) wf = = If 6 kg m 2 If = 6 kg m2; wo = ? 2 wf = 200 rpm Example The figure below shows two masses held together by a thread on a rod that is rotating about its center with angular velocity, ω. If the thread breaks, what happens to the system's (a) angular momentum and (b) angular speed. (Increase, decrease or remains the same) Examples of the Conservation of Angular Momentum Natalia Kanounnikova World Record Spin - 308 RPM Diving 18 meter high dive Amazing Fouette Turns on Pointe Playground Physics Summary – Rotational Analogies Quantity Linear Rotational Displacement Displacement x Radians q Inertia Mass (kg) I (kgm2) Force Newtons N Torque N·m Velocity v “ m/s ” w Rad/s Acceleration a “ m/s2 ” Rad/s2 Momentum mv (kg m/s) Iw (kgm2rad/s) Analogous Formulas Linear Motion Rotational Motion F = ma K = ½mv2 Work = Fx t = I K = ½Iw2 Work = tq Power = Fv Power = Iw Fx = ½mvf2 - ½mvo2 tq = ½Iwf2 - ½Iwo2 Summary of Formulas: K = Iw 1 2 Work = tq 2 tq = ½ Iw ½ Iw 2 f Height? mgho Rotation? ½Iwo2 velocity? ½mvo2 2 0 = I = SmR2 I ow o = I f w f Power = tq t = tw mghf Height? ½Iwf2 Rotation? ½mvf2 velocity? CONCLUSION: Angular Motion